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I did the following calculation, first I take the $k$-th derivative with respect $s$ of the Mellin transform of the fractional part $\{\frac{1}{t}\}$ defined for $0<\Re s<1$ as it is showed in the identity $(11)$ (see [1]) of the MathWorld's article Riemann Zeta Function. And after I equate the result to zero, to get from the RHS an equation in terms of the Riemann zeta function $\zeta(s)$ and its first few derivatives.

For the first derivate, the equation that I'evoked is $$s\zeta'(s)-\zeta(s)=0,\tag{1}$$ as was said we consider such equation for $0<\Re s<1$, where thus $s=x+iy$ denotes the complex variable.

We denote the set of double zeros of the Riemann zeta function, if there are any for $0<\Re s<1$ as $\mathcal{D}$. I've added this set as budget only in the case that your solution needs to invoke the definition of such set $\mathcal{D}$.

Question. Is it known the solution of $$\left. \begin{array}{l} &s\zeta'(s)-\zeta(s)=0\\ &0<\Re s<1, \Im s>0 \end{array} \right\}\tag{P}$$ in terms of certain solution set and/or $\mathcal{D}$? If it is in the literature refer it and I try to search and read it from the literature. In other case, is it possible to solves the problem $(P)$? Many thanks.

In short, I'm asking how a mathematician solves it and what is the meaning of your solutions. If our the problem $(P)$ makes sense for a different infinite strip $S'\subset\{s:0<\Re s<1\}$ and $\Im s>0$ feel free to discuss it, instead my problem ($0<\Re s<1$ and $\Im s>0$).

I have no intution what are the solutions of $(P)$, my only ideas should be choose a suitable representation of $\zeta(s)$ and $\zeta'(s)$ for $0<\Re s<1$, after take the real part of $(1)$, and respectively the imaginary part, maybe to use the functional equation and finally try to solve the system for some rectangle using a CAS, to get some conjecture.

References:

[1] Balazard, M. and Saias, E. The Nyman-Beurling Equivalent Form for the Riemann Hypothesis, Expos. Math. 18, 131-138, (2000).

[2] Are any zeros of Riemann zeta function and the zeros of the derivatives of Riemann zeta function same?, Mathematics Stack Exchange (September, 2015).

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    $\begingroup$ For $\Im(s)$ large enough the only roots on $\Re(s) = 1/2$ are at the double zeros of $\zeta$ because $\frac{\zeta'(s)}{\zeta(s)} -\frac12 \log \pi + \frac12 \frac{ \Gamma'(s/2)}{\Gamma(s/2)}$ is purely imaginary while $\frac1s-\frac12 \log \pi + \frac12 \frac{ \Gamma'(s/2)}{\Gamma(s/2)} =\frac1s-\frac12 \log \pi + \frac12 (\log s +O(\frac1s)) = \log |s|+O(1)$ is not $\endgroup$
    – reuns
    Jul 22, 2019 at 15:57
  • $\begingroup$ Many thanks for your remarks always @reuns , today I tried to think other symbolic calculations, I provide the formulas as a present for your patience, if you want to study it in your home $$\zeta(3)=\frac{3}{2}\int_0^\infty x^{3}\left(\sum_{n=1}^\infty n\operatorname{Ai}(n x)\right)dx,$$ where $\operatorname{Ai}(x)$ is the Airy function, and for integers $m\geq 1$ $$\zeta(3m+1)=\frac{3^m\cdot\Gamma(m)}{\Gamma(3m)}\int_0^\infty x^{3m}\left(\sum_{n=1}^\infty \operatorname{Ai}(n x)\right)dx.$$ It is from easy calculations from corresponding Mellin transform, I hope that there aren't mistakes! $\endgroup$
    – user142929
    Jul 22, 2019 at 20:22
  • $\begingroup$ $Ai(x)$ has nothing special in there $\endgroup$
    – reuns
    Jul 22, 2019 at 21:44
  • $\begingroup$ Then I suppose you are saying that the formulas are correct, and that you know how to simplify their integrands (I had doubts about it), maybe in terms of trigonometric functions @reuns $\endgroup$
    – user142929
    Jul 22, 2019 at 22:16

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